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Question Number 140652 by mohammad17 last updated on 10/May/21

Answered by TheSupreme last updated on 10/May/21

x^3 −x^2 +x−1=(x−1)(x^2 +1) ∫((3−x)/(x^3 −x^2 +x−1))dx=∫(A/(x−1))+((Bx+C)/(x^2 +1))dx Ax^2 +A+Bx^2 −Bx+Cx−C=3−x A+B=0 −B+C=−1 A−C=3 A=1 , B=−1, C=−2 ∫(1/(x−1))dx−∫((x+2)/(x^2 +1))dx=ln∣x−1∣−(1/2)ln(x^2 +1)−2tan^(−1) (x)+c ∫(1/(x(x+1)^2 ))dx=∫(A/x)+((B(x+1)+C)/((x+1)^2 ))dx Ax^2 +2Ax+A+Bx^2 +Bx+Cx=1 A=1 2A+B+C=0 A+B=0 A=1, B=−1, C=−1 ∫(1/x)dx−∫(1/((x+1)))dx−∫(1/((x+1)^2 ))dx=ln∣x∣−ln∣x+1∣+(1/(x+1))+c

x3−x2+x−1=(x−1)(x2+1)∫3−xx3−x2+x−1dx=∫Ax−1+Bx+Cx2+1dxAx2+A+Bx2−Bx+Cx−C=3−xA+B=0−B+C=−1A−C=3A=1,B=−1,C=−2∫1x−1dx−∫x+2x2+1dx=ln∣x−1∣−12ln(x2+1)−2tan−1(x)+c∫1x(x+1)2dx=∫Ax+B(x+1)+C(x+1)2dxAx2+2Ax+A+Bx2+Bx+Cx=1A=12A+B+C=0A+B=0A=1,B=−1,C=−1∫1xdx−∫1(x+1)dx−∫1(x+1)2dx=ln∣x∣−ln∣x+1∣+1x+1+c

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